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Let $u_0,u_1,u_2,...$ be a sequence of integers given by $u_0=2, u_1=3$ and $u_{n+1}=3u_n-2u_{n-1}$ for every integer $n\ge1$. Then $u_n=2^n+1$ for all $n\in \Bbb N_0$.

I let $P(n)$ be $u_{n+1}$ and then let $P(1)=u_{1+1}=3u_1-2u_0=5$ after substituting in the corresponding values. I'm confused as to what my next step should be as I'm not sure what $u_{1+1}=5$ proves.

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(Just adding some more detail about how induction works. Basically J.W. Tanner's explanation fleshed out a bit more).

Don't have $P(n)=u_{n+1}$ or something like that. Have $P(n)$ be a statement. In this case:

$P(n)$ is the statement "$u_{n}=2^n+1$"

Base Case

$P(0)$ is true since $u_0=2=2^0+1$

$P(1)$ is true since $u_1=3=2^1+1$

Induction Step

Now, using (strong) induction: assume $P(0), ..., P(n)$ and we will prove $P(n+1)$.

$u_{n+1}=3u_n-2u_{n-1}$ by definition. Let's subsitute in the values for $u_n$ and $u_{n-1}$

$u_{n+1}=3(2^n+1)-2(2^{n-1}+1)=3\times2^n+3-2^n-2=2\times2^n+1=2^{n+1}+1$

Which verifies that $u_{n+1}=2^{n+1}+1$ which verifies $P(n+1)$.

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$u_0=2=2^0+1$ and $u_1=3=2^1+1$.

Now assume $u_{k-1}=2^{k-1}+1$ and $u_{k}=2^{k}+1$. Then

$u_{k+1}=3u_{k}-2u_{k-1}=3\cdot2^{k}+3-2\cdot2^{k-1}-2=3\cdot2^k-2^k+1$

$=2\cdot2^k+1=2^{k+1}+1$.

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